Where Does y=x²+2x-8 Become Positive? Solve the Quadratic Inequality

Look at the following function:

$y=x^2+2x-8$

Determine for which values of $x$ the following is true:

$f(x) > 0$

To solve this problem, we will perform the following steps:

• Step 1: Find the roots of the quadratic equation $x^2 + 2x - 8 = 0$.
• Step 2: Use the roots to determine the intervals where the quadratic function is greater than zero.

Step 1: We start by finding the roots of the equation $x^2 + 2x - 8 = 0$.
We use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 2$, and $c = -8$.

Substitute these values into the formula:
$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}$.

This gives us two roots:
$x_1 = \frac{-2 + 6}{2} = 2$
$x_2 = \frac{-2 - 6}{2} = -4$

Step 2: Use these roots to determine intervals on the number line: $x < -4$, $-4 < x < 2$, and $x > 2$. Since the parabola opens upwards (positive coefficient of $x^2$), it is negative between the roots and positive outside of them.

Hence the solution to $x^2 + 2x - 8 > 0$ is the combined intervals $x < -4$ or $x > 2$.

Therefore, the solution to the problem is $x > 2$ or $x < -4$.